Proving Quadrilateral is a Parallelogram

Date: 11/30/2001 at 12:24:10
From: Kara
Subject: Proving Quadrilateral is a parallelogram
We are having a problem with the idea of a quadrilateral having one
pair of opposite sides congruent and one pair of opposite angles
congruent. We've flipped the figure and found isosceles triangles, so
we THINK that we can say a quadrilateral is a parallelogram if these
conditions exist.
Is there a theorem that states you only need one pair of opposite sides
and angles congruent for it to be a parallelogram?
If there isn't, can you just give us a clue to how we can disprove it?

Date: 11/30/2001 at 16:44:04
From: Doctor Rob
Subject: Re: Proving Quadrilateral is a parallelogram
Thanks for writing to Ask Dr. Math, Kara.
The statement you want to be true isn't. Here is a way to construct a
counterexample.
Draw an acute angle <A. From some point B on one side, using a large
enough radius, draw an arc of a circle that intersects the other side
in two points C and D:
B _,-'
_,o'
_,-' / \
_,-' / \
_,-' / \
_,-' / \
A o--------------o---------o-----
D C
By construction BC = BD, since they are radii of the same circle.
Notice that AC > AD. Now make a copy EFG of triangle ABD, and erase
line segment BD and point D:
B
_,o
_,-' \
_,-' \
_,-' \
_,-' \
A o------------------------o C
F
_,o
_,-' /
_,-' /
_,-' /
_,-' /
E o--------------o
G
Rotate it around so that F coincides with C and G coincides with B.
This is possible because FG = BD = BC = CB. Then you have a
quadrilateral like this:
E
o
/|
/ |
/ |
/ |
/ |
B=G / |
_,o |
_,-' \ |
_,-' \ |
_,-' \ |
_,-' \|
A o------------------------o
C=F
Now erase line segment BC, and you have a quadrilateral ACEB.
E
o
/|
/ |
/ |
/ |
/ |
B / |
_,o |
_,-' |
_,-' |
_,-' |
_,-' |
A o------------------------o
C
By construction EC = EF = AB, and <A = <E, so two opposite sides are
equal and two opposite angles are equal, but this is definitely NOT a
parallelogram, because EB = EG = AD < AC.
You may object that the above quadrilateral is not convex, that is,
it has an interior angle greater than 180 degrees. In this
counterexample, that is true. If <A had been chosen larger, and points
C and D closer together, a convex quadrilateral satisfying your
conditions, but still not a parallelogram, would have resulted.
Feel free to write again if I can help further.
- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/